Let the orthocentre and centroid of a triangle be A(-3, 5) and B(3, 3) respectively. If C is the cir

Q: Wait, why is the ratio $2:1$ and not $1:2$? I always get confused about which point is which.

That is a fantastic question! It is a common point of confusion. Think of it this way: the Centroid is the 'center of mass' of the triangle, and it is always closer to the Circumcentre than to the Orthocentre. The order on the Euler Line is $A$ (Orthocentre) - $B$ (Centroid) - $C$ (Circumcentre). Since $B$ is between $A$ and $C$, and it is closer to $C$, the segment $AB$ is longer than $BC$. Specifically, $AB$ is twice the length of $BC$, so the ratio $AB:BC$ is $2:1$.

Q: Can I use the property $3G = O + 2C$ instead of the section formula?

Absolutely! That is a brilliant observation. The property $3G = O + 2C$ (where $G$ is the centroid, $O$ is the orthocentre, and $C$ is the circumcentre) is just a rearranged version of the section formula. If you prefer working with vectors or this specific algebraic form, it will lead you to the exact same result. Just be careful with your signs when substituting the coordinates!

Q: Why do we need the distance formula at the end? Can't we just stop after finding $C$?

I understand the temptation to stop once you have found the coordinates of $C(6, 2)$! However, always re-read the final sentence of the question. It asks for the 'radius of the circle having line segment $AC$ as diameter.' Finding $C$ is only the halfway point. You must now find the length of the diameter $AC$ using the distance formula, and then divide that length by $2$ to get the radius.

Question

Let the orthocentre and centroid of a triangle be $A(-3, 5)$ and $B(3, 3)$ respectively. If $C$ is the circumcentre of this triangle, then the radius of the circle having line segment $AC$ as diameter, is :

Options:

(A) $\frac{3\sqrt{5}}{2}$

(B) $\sqrt{10}$

(C) 2\sqrt{10}

(D) 3\sqrt{\frac{5}{2}}

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the coordinate plane, looking at a triangle. You are given two points: the Orthocentre $A(-3, 5)$ and the Centroid $B(3, 3)$.

In the world of geometry, these points are part of a secret, elegant structure known as the **Euler Line**. This line is a fundamental property of all triangles, connecting the Orthocentre, the Centroid, and the Circumcentre.

The beauty of this problem lies in recognizing that these three points are not just floating in space; they are perfectly aligned.

### The Balancing Act

The key to solving this problem is the relationship between these points. The Centroid $B$ acts as a bridge, dividing the line segment connecting the Orthocentre $A$ and the Circumcentre $C$ in a strict $2:1$ ratio.

This means that if we walk from $A$ to $C$, the Centroid $B$ is positioned such that the distance $AB$ is twice the distance $BC$. Mathematically, we express this using the section formula.

For the $x$-coordinate, we have:

$$x_B = \frac{2x_C + 1x_A}{2 + 1}$$

Substituting our known values, we get:

$$3 = \frac{2x + 1(-3)}{3}$$

By multiplying both sides by $3$, we get $9 = 2x - 3$, which simplifies beautifully to $2x = 12$, giving us $x = 6$.

We perform the same dance for the $y$-coordinate:

$$y_B = \frac{2y_C + 1y_A}{2 + 1}$$

Substituting the values, we get:

$$3 = \frac{2y + 1(5)}{3}$$

Again, multiplying by $3$ gives $9 = 2y + 5$, which leads to $2y = 4$, so $y = 2$. We have successfully uncovered the coordinates of the Circumcentre $C(6, 2)$.

### The Final Stretch

Now that we have the coordinates of $C(6, 2)$, we are almost at the finish line. The question asks for the radius of a circle where the line segment $AC$ is the diameter.

First, we calculate the length of the diameter $AC$ using the distance formula:

$$AC = \sqrt{(6 - (-3))^2 + (2 - 5)^2}$$

This simplifies to:

$$AC = \sqrt{9^2 + (-3)^2} = \sqrt{81 + 9} = \sqrt{90}$$

We can simplify this radical to $AC = 3\sqrt{10}$. Finally, the radius $R$ is simply half of the diameter:

$$R = \frac{3\sqrt{10}}{2}$$

To match our answer with the provided options, we can bring the $2$ inside the square root:

$$R = 3\sqrt{\frac{10}{4}} = 3\sqrt{\frac{5}{2}}$$

And there it is! The elegance of the geometry leads us directly to the **final answer of $R = 3\sqrt{\frac{5}{2}}$**. Remember, every complex problem is just a series of simple, beautiful steps waiting to be discovered.

Animated Solution for Mathematics - Straight Lines: Let the orthocentre and centroid of a triangle be $A (- 3, 5)$ and $B (3, 3)$ respectively. If $C$ is the circumcentre of this triangle, then the radius of the circle having line segment $A C$ as diameter, is :

Select Answer:

Visualized Solution

Visualizing the Given Points

The Euler Line Property

Section Formula for $x$ -coordinate

Solving for $x$

Section Formula for $y$ -coordinate

Solving for $y$

Coordinates of Circumcentre $C$

Distance Formula for Diameter $A C$

Calculating the Length of $A C$

Finding the Radius

Matching with the Options

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Balancing Act

The Final Stretch

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Select Answer:

Visualized Solution

Visualizing the Given Points

The Euler Line Property

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Section Formula for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Coordinates of Circumcentre .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Distance Formula for Diameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Calculating the Length of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC